1. science
2. mathematics
3. algebra

PERMUTABLE POWER SERIES AND
REGULAR ITERATION
I. N. BAKER
(received 3 August 1960)
1. Introduction
If f(z) is a power series
(i)
n-l
convergent for \z\ < p, where p > 0, then f(z) is said to have a fixpoint of
multiplier ax at 2 = 0. In the (local) iteration of/(?) one studies the sequence
{/„(*)},» = 0, 1, 2, • • • in a neighbourhood of z — 0, /n(z) being defined by
/oto = * . / « « = / 0 - i t o }
^r » = 1,2,3,...
For many values of the multiplier ax, including 0 < \ax\ < 1 and \ax\ > 1,
the local iteration of f(z) is completely mastered by the introduction of
Schroder's functional equation
(S)
Koenigs [11] showed that, if 0 < l^l < 1 or if \ax\ > 1, there is a unique
function x(z) which satisfies (5) and has an expansion
n-2
convergent in some neighbourhood of z = 0. Any other solution of (S) which
is regular at z = 0 is a constant multiple of %{z)- Not only can one express
all the "natural" iterates fn{z) as
in suitable neighbourhoods of z = 0, but with arbitrary real or complex X
and suitable determinations of ax the expression
(2)
AW = z-i{«iA*(*)}
generalises fn{z) to a family of continuous iterates fx(z) which are permutable
in the sense
265
266
I. N. Baker
[2]
(3)
/*{/,(*)} = /,{/*(*)} = /*+,(*)•
The same calculations may be carried out formally in the case ax = exp(t'0)
where 0 is real and irrational. Cremer [4] and Siegel [14] have shown that
the convergence of the series obtained depends on number-theoretic properties of 0. If «! is a root of unity, even the formal calculation of the Schroder
function is generally impossible and if ax = 1 this is always the case.
This paper is concerned entirely with the case ax = 1. The natural iterates
fn[z) have been studied by Fatou [5] and others, but continuous iterates
analytic at z = 0 are not provided by the above Schroder method. There is,
however, another approach based on the connexion between permutable
functions and iterates (c.f. Hadamard [7], Szekeres [15]). We use the symbol
fog, where / = 2£-i anz" a n d g = 2 X i Kz"- a r e formal power series, to
denote the formal power series 2£LiCnzn obtained by substituting g for z
in / and rearranging according to powers of z. Since cB depends only on
alt • • -an and &i, • • • &« there are no problems of convergence and the
operation fog is associative. If / is a given series with ax = 1 we may
determine all series of the form g which are permutable with / in the sense
(4)
fog = gof.
This is done in §2 and the set of such g is found to include a family F of
series fx
F : /A = z+Xam+1zm+* + £
bn{X)z\ m ^ 1, am+1 ^ 0
n+2
which satisfy
/A ° //. = //. ° fx
and may be regarded formally as a family of continuous interates. The
discussion of the convergence of the series /A in a given case is likely to be
difficult since the rule of formation of the bn(X) is complicated. In [1] the
special case corresponding to fx = ez—1 was treated by indirect arguments
to show that in that case only integral values of X yield convergent series
(at least for real X). This prompts the question: what A-sets can arise as the
sets of those values A for which the member /A of a family (F) has a positive
radius of convergence? The answer is provided by § 6, theorem 10: such a
set consists of the whole A-plane or of one or two dimensional lattice of
points. In the first case f(z) is embedded in a continuous group F of iterates
which are regular at z = 0 and which possess a regular infinitesimal transformation (§ 7). In the other cases no such embedding is possible.
Sections 3,4 contain extensions of the work of Fatou [5] which are used
in the proof of theorem 10. Section 5 contains a proof that those /A of a
family which have positive radius of convergence satisfy certain conditions
[3]
Permutable power series and regular iteration
267
of boundedness (theorem 7) which are essential for the later proofs.
Besides studying iteration and permutability in the neighbourhood of a
fixpoint one may consider e.g. permutable rational functions (Julia [10]),
polynomials (Ritt [12] and Jacobsthal [9]) or entire functions (Baker [1]
and Ganapathy Iyer [8]). In § 8 some applications of our results to such
global problems are considered.
2. The formal power series
The extent of the formal power series which are permutable with a given
formal power series and the way in which these series form a continuous
family are described in the following theorems.
THEOREM
(5)
1. If f(z) is any formal series of the form
/(*) = * + I
anz", am+1 # 0, m ^ 1,
n-m+l
then every formal series
(6)
g{z)=Zbnz*
»=i
which is permutable with f(z) in the sense of (4), has J™ = 1.
If bk, k ^ 2 is the first nonzero coefficient in (6), then the comparison of coefficients of powers of zm+k in (4) gives kam+1bk = 0, which is a
contradiction. Thus only the case k = 1, bx ^ 0 is possible and the comparison of coefficients of zm+1 in (4) gives am+1(bt — b™+1) = 0, so that 6J* = 1.
PROOF:
THEOREM 2. Case bx = 1. With the notation of theorem 1: if bx = 1,
then b2 = • • • = bm = 0 while bm+1 is arbitrary and to an arbitrary value X
there is exactly one g(z) with bm+1 = ham+1. This g[z) we denote by fx.{z) and it
has the form
(7)
h(z) = z+Zam+1z^
+ | bn[X)z*.
n—m+2
where bn(X) is a polynomial in X of degree at most (n — m).
Equating coefficients of zm+2 in (4) gives (m — l)am+1b2 = 0 and
thus b2 = 0 if m > 1. Equating successively coefficients of zm+i for / = 3,
• • • ,m, gives (m — j + l)am+1bt — 0 and thus 63- = 0. For z2m+1 the coefficients vanish identically, while for k > m + 1 coefficients of zm+k give an
equation of the form
PROOF:
(8)
<Wi(»* + 1 ~ k)bk = *Z Cibt + X d^.^b
3-m+l
268
I. N. Baker
[4]
where the non-negative integers p,q,- • 'V satisfy
(9)
m+l<p
+ q-\
\-v ^m
+ k~
1
and
(10)
p+
{m+l)q+'--(k-l)v
= m + k,
while the ct and &va...v are dependent only on the coefficients an. It follows
from (8) that bm+1 is arbitrary and that once bm+1 is given all other bn,
n > tn + 1 are determined inductively in a unique way. If we set bm+l =
Xam+1 and calculate the bn they will be polynomials in X with coefficients
depending only on the an. The degree of bn is at most n — tn, as is proved by
the following induction: for n = tn + 1 the degree is 1. Suppose for n < k
the coefficient bn has degree at most n — tn in X. In equation (8) every term
of the first sum on the right has degree at most [k — 1)— tn <k — tn in X
while the degree of any term of the second sum is (using (10))
D ^ q + 2r ~\
(- {k - 1 - m)v
= p+(m+l)q+
— p — tn{q + r -\
[m + 2)r-\
h v}
+ (k - l)v
= tn -\- k — p — tn{q -\- r -\- • • • -\- v).
Iiq + r + --- + v=l then p7>tn from (9) so that D <Lk — tn, while if
q + r + ••• + v^2 then D is at most m + k — 2tn = k — tn. Thus the
right hand side of (8) has degree at most (k — tn) and the same is true of bk.
We note some useful consequences of theorem 2 in
THEOREM 3. The series denoted in theorem 2 by /A (z) coincide with the natural
iterates in the case when X is a positive integer. For any two values X and /i one
has
PROOF: The natural iterate fn(z) defined by / o / o • • • o / with n factors is
a series of the form
which is permutable with f(z). By theorem 2 there is only one series with this
property, namely the series fx(z) with X — n. The two are identical.
In the same way, since /A o fM o f = fx o / o /,, = /A o /,, o / and /,, o fK o f =
f ofpO fx, while both /A o /^ and /^ o fK have the form
z +
{X + f i ) a m + 1 z m + 1 +
---,
equation (11) follows.
Combining equation (11) with theorem 2 one sees that the /A(z) form a
[6]
Permutable power series and regular iteration
269
continuous group of formal power series, all of which are determined by any
one member other than fo(z) which is the "identity" z. Thus each of the
coefficients bn(X) has the properties
(12)
6.(0) = 0, bh(\) = an,n^m+l.
.
We now turn to the case where b™ = 1 but bx ^ I. This arises only when
m > 1. We make a preliminary change of variables.
LEMMA
1. There is a formal power series
(13)
w = P(z) = z + X2z* + V
3
+ ••'
such that if f(z) is given by (5) with m > 1, then P o / o P_x = F(w) has
the form
F(w) = w + Aw^1 + Bit)*"*1, A =£0.
(14)
PROOF: If m > 1 we compare coefficients in P o f = F o P and try to
determine the Xn so as to give F the form (14). Coefficients of z* for k = 1,2,
• • • m yield nothing, while zm+1 yields A = am+1 and zm+k for k = 2, • • • m
yields an equation of the form
(m + 1 — k)am+xXh = E(X2, • • • At_1(aa ,
m+1
Thus A* is determined for k = 2, • • • m. Equating coefficients of z2m+1
gives an equation which does not involve Xm+1 but which determines B in
terms of the a's and A2, • • •, Am. Xm+1 may be given an arbitrary value.
Equating coefficients of zm+k for k = m + 2, m + 3, • • • yields an equation
(tn+l-
k)am+1Xk =
which determines all further Xk.
THEOREM 4. For f(z) as in (5) theorem 1, tffore is to each value of bx which
satisfies b™ = 1 exactly one formal power series
(15)
H{z)=J.bnz»
n=l
/or which Hm{z) = 2 teiAi/e H o f = f o H. The series g(z) of the form (6)
which are permutable with f(z) are exactly the series H o / A , where /A Aas the
meaning described in theorem 2.
PROOF: Let bx be any root of &
™ = 1 and let P(z) be any power series
(13) satisfying the conditions of Lemma 1. If F = P o f o P_x and
g(z) = bxz + . . . is permutable with f{z), then G = P o g o P_1 = bxz -\- . . .
is permutable with F. Since .F(z) has the form (14) it is permutable with
the series btz and its [m — 1 )-th iterate b^z, and hence with (b^z) o G=J^ 1 G.
270
I. N. Baker
[6]
Since b^P o g o P_x commutes with F, it follows that k = P_x o bxxP o g
o P_x o P = P_i° i j 1 P o ^ commutes with /. Now k = z + . . . and by
theorem 2 must be identical with one of the /A. Hence we have
g = P_x o bxP o /A = H o fx where # == P_x o 6X P. Clearly Hm=P^obfP = z.
AlsoHof= P_ 1 o6 1 PofoP_ 1 oP = P_ 1 o6 1 FoP= P_x o .F o (bxz) oP = P_x
oFo P_x obx P = f o H. It remains to prove the uniqueness of H = bxz + . . ..
Suppose K = bxz -\- . . . also commutes with / and satisfies Km = z. Then
H oK_x commutes with / and has an expansion z + • . ., so that it must
be identical with some /A. We shall prove in theorem 5 that H and K commute so that
{HoK_1)n = U={HmoK_m)^z
and A = 0.
THEOREM 5. Any two seriesg(z) andh(z), each of the form (6) and permutable
with f(z) of the form (5), satisfy g o h = h o g.
PROOF: If g and h both have the form fx the result follows from theorem 3.
If h = H(z) is of the form (15) and if g = /A then by the uniqueness result
of theorem 2 there is a JU such that
(16)
# o / A o #_! = /„.
If H and H_1 have expansions
n-l
then the comparison of terms in zm+1 in (16), using (7) and the fact that
H and H_x are formal inverses, gives
so that H o /A = /A o H.
Further, if H, K are two solutions of Hm = Km = z which are permutable
with /, then H o K = bz + • • • where bm = 1. Thus (H o K)m = z + higher
terms and is permutable with / so that from theorem 2 there is a A such that
(HoK)m = fx. But then (K o H)m = K o (H oK)m oK_x = K o /A o tf^
= /A since by the preceding part of the proof K is permutable with /A. Now
because its expansion has 1 for the coefficient of z, H o K o H_x o i ? ^ must
be fa for some a. Then (H o i<Qm = (faoKo H)m = fma o (K o H)m since / a
is permutable with both H and if. This means that A = mx + X so that« = 0,
fa = z and H oK = K o H. This is the result which was needed in the last
step of theorem 4, which is now completely proved. Using the result of
theorem 4 it now follows that in every case the statement of theorem 5 holds.
[7]
Permutable power series and regular iteration
271
It may be noted that the result of theorem 5 depends essentially on the
restriction to series having a common fixpoint (namely z — 0). It is not
always true, for example, that if the entire functionsg(z), h(z) are permutable
with the entire function f{z), then g{z) is permutable with h[z). A counter
example is provided by the functions
f(z) — z + sin z + sin(2 + sin z),
g(z) = — z — sinz,
h(z) = 2n — z — sin z,
for which g2(z) = hz(z) = f(z). In fact g(z) and h(z) have no common fixpoint and g(h) — h(g) = — 4n. The purely formal considerations of this
section therefore have consequences for such global problems as the set of
entire functions which commute with a given entire function F(z). As an
example we prove:
THEOREM 6. / / F(z) is an entire function exactly one of whose fixpoints has
multiplier 1, then any two non-constant entire functions permutable with F{z)
are permutable with one another.
PROOF: We may suppose the fixpoint in question to be z = 0 for if F(z)
has fixpoint c with F'(c) — 1, then f{z) = F(z + c) — c has fixpoint z = 0
with multiplier /'(0) = 1. Also if F(G{z)) = G(F(z)) a.ndiig{z) = G(z+c) -c,
then f{g(z)) = g(f(z)) and conversely.
If G(z) is any non-constant entire function satisfying F(G(z)) = G(F(z)),
it follows from .F(0) = 0 that G(0) is a fixpoint of F(z). If one differentiates
the identity F(G(z)) = G(F(z)) and sets z = 0 one obtains F'(G(0))G'(0) =
G'(0)F'(0) = G'(0), so that .F'(G(0)) = 1 unless G'(0) = 0. Since G(z) is not
a constant there is a least value of m such that the derivative G(m) (0) ^ 0
and w-fold differentiation of the above identity yields for z— 0 the equation
so that F'(G(0)) = 1 and F(G{0)) == G(0). By hypothesis the only fixpoint
of multiplier 1 of F(z) is z = 0, and hence G(0) = 0. The expansion of G(z)
in powers of z has no constant term and is therefore contained among the
formal series of the form (6) satisfying G o F = F o G. Since any two of
these series are formally permutable, any two representing entire functions
will be permutable as entire functions.
We shall return to such global applications after considering the convergence of the formal series obtained in theorems 1 to 5.
3. Natural iteration near a fixpoint of multiplier 1
The results of this section are mainly due to Fatou [5] but are developed
in forms suitable to the applications made of them in § 4. It is convenient
272
I. N. Baker
[8]
to put the fixpoint at infinity and to begin by treating a function defined
for all sufficiently large values of z by the convergent series
(17)
F(z)=z
+ a + - + ^-+---,x>
Z
l,«>0.
IT
The expansion in descending powers of z may contain fractional powers which
can be assumed to be made single valued by cutting the region of convergence
along the negative real axis. In fact we assume all the exponents of z in (17)
to be multiples of l/« for some integer n > 1. Now to a given value 6 with
0 < 0 < n/2 there is a positive constant K such that for \z\ ^ K the series
(17) converges and
(18)
\F(z) —z — a\<asin6
for \z] ^ K .
We denote by (£ the curve formed by the circular arc
Fig. 1.
together with the two infinite segments of the tangents at the extremities
of this arc which extend from the extremities to infinity in the left half plane.
The region ® = 55 (6, K) is defined to be the region bounded on the left by (5.
LEMMA 2. / / F(z) has the form (17), 6 satisfies 0 < 6 < n\2 and 2) =
% (0, K) is defined as above, then for all sufficiently large K
(19)
* • „ ( * ) € S>, » =
1,2,.-•
and
(20)
Re Fn{z) -> oo as n -> oo
for all z in the closure ® of 2).
PROOF: From (18) follows (19) in the case » = 1 and by inductive iteration of this result the other cases also follow. From (18):
[9]
Permutable power series and regular iteration
Re{F(z)
— z}>a(l
— sin0)
273
for z « J
and since Fm(z) e 2) for m, = 1, 2, • • • one has
Re{Fm+1(z) - -F»(*)} > «(1 - sin 0) for * e ©J m = 0, 1, -. • •, » - 1.
Adding these inequalities one obtains
(21)
Re{Fn(z) — z}> »(1 — sin 0) for * e 5^ » = 1, 2, • • •
from which (20) follows.
By a straightforward calculation Fatou [5] shows that if 3) = ®(0, if)
is a region satisfying lemma 2, then the limit
(22)
A(z) = lim \Fn(z) - na
log n]
exists uniformly for z e 3). The function .4 (z) is regular in ®. If K is sufficiently large F(z) is schlicht in 2) and so consequently are all Fn(z), n = 1,
2, • • • It follows that A (z) is either schlicht in $ or a constant, but from (22)
and (17) one sees that A (z) satisfies
(23)
and is not a constant but a schlicht function.
LEMMA 3. If F{z) has the form (17) and 9 satisfies 0 < 6 < nj2, then for
all sufficiently large K the domain % = 2)(0, K) satisfying lemma 2 has the
further property that the function A (z) satisfying (22) and (23) is schlicht in 2)
and A'(z) -*• 1 uniformly « s : - > o o i » 5).
PROOF: We have only to prove that A'(z) ->• 1 as z -*• oo in ®. One may
assume K to be so large that for all z e ®
where B > Ois a suitable contant; and if can bs enlarged further if necessary so that B < K2. Then
(25)
A < i
for
*e®.
Now
and if z — z0 e 3), then for / = 1, 2, • • • we have z, = Ff{z) e © and
274
I. N. Baker
(26)
[10]
n (i-si^n < \K(*)\ < n (i + B\z,n.
i-0
i-0
But from (22) A'{z) = lim^^F^z) and hence
(27)
ft
(1 - B\zt\-+) ^ \A'(z)\ ^ ft (1 + B\z,\-*).
Enlarging K still further will not alter the validity of the above arguments
for ze ®(0, K) = % and we suppose K so large that for \z\ >K and in
particular for z e 2) one has
(28)
\F{z) - z - a\ < a sin 0/2.
For any z0 e ® the points zn = i%,(z0) lie in an angle
(29)
|arg(* - , 0 )| < 0/2
whose vertex is at z0 and whose bisector extends in the direction of the
positive real axis. Thus if
(30)
|arg* o |^(rc-0)/2, z0 e %
then
(»1)
I'ol < I'll < ' ' ' < I*J < l-Zn+ll < ' ' *
If on the other hand
(32)
|arg zo\ > (n - 0)/2, z0 e ®,
then the radius vector from 0 to z0 makes with the nearer arm of the angle
(29) an angle y lying between 0/2 and n/2. Since all zn lie in the angle (29)
and the nearest point of (29) to 0 has modulus \zo\ sin y it follows that in
this case (32) one has
(33)
|*J^|* 0 |siny>|* 0 |sinfl/2
Thus for any z0 e $ we have either (30) or (32) and hence either (31) or (33)
so that in any case
(34)
\zn\ > \zo\ sin 0/2 = \zo\t for n = 1, 2, • • •
where t = sin 0/2. From (21) with s = 1 — sin 0 one has
(35)
\Fn{z0)\ SiRe Fn(z0) > ns - |*0| for * > |*0|/s.
For n > [2|^0|/s] + 1 = N, where the square bracket means "integral part
of 2|zo|/s", one has |zo| < sn/2 and from (35) follows
(36)
|Fn(z0)| > »s/2 for n > N and z0 e ©,
Using (34) and (36) for z0 e ® one has
[11]
Permutable power series and regular iteration
oo
N—X
n=0
0
275
oo
Q = 2 I*J-2 = 1 + 2
N
N
~
4
^2
'N
4
•<«|*o
2
2
1
(2|*0|-s)s
Thus the series Q converges for x0 e 2) and (? -> 0 uniformly as z -*• oo in ©
so that the infinite products in (27) converge. Since the left hand product
of (27) is greater than 1 — BQ while the right hand one is less than exp(5())
the lemma is established.
4. Transformation of the permutable series
A change of variables is now made to replace the series of a family (7)
by series (46) of the form (17). In the commuting family
(?)
A W = * = + ^ m + i 2 m + 1 + I KWzn
n-m+2
put
»=2
where the An are so chosen that
Pof1oP_1{w)
= Q1(w)
has the form
(38)
Qx(w) = w + am+lwm^ + f Bnw\
n=2m+l
The possibility of so doing has been shown by Fat'ou [5, §§ 11, 12]. For
m = 1 the transformation reduces to w = z. If for each value of X one sets
(39)
ft(»)
= Po/ A oP_ 1 (w),
then the Qn form a family of mutually permutable series whose expansions
begin
Qx(w) =w +
276
I. N. Baker
[12]
By direct calculation in QxoQx = Qxo Qx one finds that all Qx have the
form
(40)
Qx{w) = w + Xam+1w">+i + |
Bn{X)w\
n-2n»+l
By moving the fixpoint w = 0 to infinity and rotating axes:
(41)
w = e-'rv-1, Rx(v) = e-*~>Qj1•(«-*>'v-1), y constant,
where Qj1 is the reciprocal of @A, we obtain
(42)
RK{v) = v - Xam+1 exp(-»ym)»i-» + f
cn{X)v^-\
We choose the constant y in such a way that
(43)
a= - am+1 exp(-iym) > 0
and (42) becomes
(44)
Rx(v) = v + Xav*--m + f cB(A)i>*-".
n~2fn
Finally we set
(45)
v = <v« i?A (<) = {i?A (<iM) }«f
and obtain an expression
(46)
Fx (t) = t + Urn + f Dn (A)f-(»+")/m,
n-0
in which the Dn(X) are uniquely determined by the formal transformations
(37) — (45). Moreover, if for a value X the series fK of (7) has a positive radius
of convergence then there is a K > 0 such that for any choice of txlm in (46)
the series Fx converges for |2| ^ K, and conversely. We introduce the notation:
® = S(/A) is the set of values X for which the series fx of the family (7) or
equivalently Fx of (46) have positive radius of convergence.
Further, the notation
(47)
Tt = t)'m = p">, j = 1, 2, • • •, m
is introduced for that branch of the function f-lm for which
(48)
(2/ - 3)a/m < arg Tf ^ (2/ - l)»/m.
For X c ®, 1 ^ / ^ »t the substitution of (47) in (46) gives a function
(49)
F{(t) = t + Xam + £
n-0
[13]
Permutable power series and regular iteration
277
regular and single valued in neighbourhood of t = oo cut along the negative
real axis.
For 0 < 0 < n/2 and all sufficiently large K the domain 3) (0, K) in the
2-plane is mapped by the composition of the transformations (37), (41),
(45) — (47) one to one and conformally on to a domain © (j, 0, K) of the zplane bounded by a simple closed curve analytic except at z = 0 where two
boundary arcs meet in the directions
V
;
|arg x = - y + 0/m - (2/ - \)n\m,
(arg z = — y — d/m — (2/ — 3)n/m.
For K > L one has $(0, K) C 25(0, L) and hence
&(j,6,K) C%{j,d,L).
By taking K sufficiently large one may make © (/', 0, if) interior to any given
circle \z\ < p, p > 0. For large if the m domains © (/, 0, K) exist and are
ranged about z = 0 in a star @(0,if):
(51)
Figure 2 illustrates <5(d,K) in the case w = 3, y = 0
LEMMA 4. / / /A a « i F\, 1 fg / fS m, are members of the families (7) and
(49) respectively, if A e f ««<f 0 < 0 < ^/2, £Ae» /or «// sufficiently large K
one has
(52)
^i{$(0, K + 2\X\am)} C ®(0, K),
while if X > 0
(53)
i
278
I. N. Baker
[14]
Further the functions fx{z) and Fx(t) represent the same transformation in z
and t coordinates respectively so that
(54)
/A{<5)(/, 6, K + 2\l\am)} C ©0'. 0, K)
and
(55)
/A{©(0, K + 2\l\am)} C ©(0, K)
hold for all sufficiently large K, while if X> 0:
fx{<S{d,K)}C<S{d,K).
PROOF: Equations (52) and (43) follow from the form of (49) and the fact
that a > 0. We now show the equivalence of the functions Fx(t) and fx{z)
for X e g . The functions T,{t) = tVn of (47) and Rx(v) = Rx(T,(t)) of (44)
are regular for t e 2)(0, K) = % if K is large enough. Also tx = F{(t) in (49)
is regular for * in ® = ®A(0, K) D ®A(0, tf + 2\X\am) = ®0, and for t e ®0
we have tx c 5) by (52). We also suppose ./£ to be so large that the series of
transformations (37), (41), (45), (47) map $ one to one and conformally on
to ©(/, 0, K) as explained immediately after equation (49). If the successive
images of t and tx under these transformations are v and vlt w and wlt
z and z± respectively, we have to show that for t e S)o and tx = F*x{t) one has
This done, (54) and (55) follow at once from (52). Now
is regular for f e 3)0 since fx e © and S) does not contain the origin. But from
(45) follows
*r = h = {Ri(T,{t))}",
whence
(56)
Vl
= i7/?A(») =
t,Rx{Tt{t))
where tf1 = 1. Since both sides of (56) are regular for t e 3)0 the same value
of r\ must be valid for all t e 3)0. But astf->• oo along the positive real axis it
follows from (49) and (47) that
arg vx = arg T^) ~ arg T(t),
while from (44)
arg Rx{v) = arg R^T^t))
~ arg T(<).
Hence j ; = 1. Thus «! = R\(v) and since the remaining transformations
(37) and (41) are one to one in the regions under discussion, it follows that
[15]
Permutable power series and regular iteration
279
A consequence of Lemma 4 is that F'nX(t) is indeed the «-th iterate of Fx(t)
and Fx(t) commutes with F^t) in suitable regions of the form 5)(0, K).
We note the following
LEMMA 5. With the assumptions of lemma 4, if X eft and l e t , then for all
sufficiently large K one has for all t e 2)(0, i£ +
(57)
K{.n{t))
= Fx(FUt)) e $(0, K), n = 0, 1, 2, • • •
For n = 0 the result is included in lemma 4, since Fj0{t) = t.
We suppose the result holds for 0, 1, • • •, n — 1. We also suppose K so large
that for all L > K equation (53) holds with X = 1:
PROOF:
Then for t c $(0, K + 2\X\am), F{t) e ®(0, K + 2\X\am) and by the induction
hypothesis
= F>n(F[(t))
5. Normality of the family fx
THEOREM 7.
(7)
Let
h{z) = z+Xam+1zm+i + |
anz\ am+1 # 0, m ^ 1,
be a commuting family of formal -power series as in theorems 1 and 2, For
p > 0 let $„ be the class of complex X with \X\ £S p for which fx(z) has positive
radius of convergence. Then there exist constants p > 0 and M > 0 such that
(i)
and
(ii)
/A (2) converges in \z\ ^p
for all Xeftj,
\f\(z)\ < M uniformly for all \z\ ^p
and all ^ « S t .
If $tp contains only X = 0 there is nothing to prove. In the other
cases there is no loss in generality if we assume X = 1 to be contained in Stj,
for there is certainly a p, ^ 0, ft e $„, and the substitutions X = fiX', fp(z) =
gi (2) < fx (z) = £A' {*) reduce the general case to the special one (with a different
value of p).
We assume then that p > 1 and that X = 1 is contained in $„. We choose
a fixed integer / satisfying 1 5S / ^ m and set
PROOF:
(58)
F(t) = F{(t) =t + ma + 2 Dn(l)T&)-«>+»
n-0
for that function (49) for which X = 1. F(t) is simply the function f^z)
280
I. N. Baker
[16]
represented in a different system of variables. Since ma > 0 by (43), the
series (58) is of the form (17) and lemma 3 applies. Thus if 0 is a fixed number
satisfying 0 < 0 < n/2, the domain ® = %)(d,K) described in § 3 will for
all sufficiently large i f b e a domain of existence and univalence of F(t) and
of the function A (t) (defined by (22) with z replaced by t). Since A '(t) -> 1
as t -*• oo in ® one may take K so large that for alltfe 2> one has
(59)
|aigil'(0|<fl/2,|il'(*)|>l>0.
These conditions ensure that s = A (t) maps 2) one to one and conformalyl
on to a region 93 of the s-plane with the property: 33 is the region lying to the
right of a simple analytic curve 2 which is cut just once by every parallel
to the real axis and whose limiting slope at infinity is — tan 0 in the upper
half-plane and + tan0 in the lower half-plane (c.f. Fig. 3). By (59) the
Fig. 3.
slope of £ has a numerical value greater than 0/2.
We denote by ^ the region formed by displacing S3 an amount p dm,
where a = a/(sin 0/2), in the direction of the positive real axis. The distance
of a point of 93x from £ is at least patn.
We now suppose X to be any fixed value in ®v. By lemma 5 there is an L
greater than the K used above in the definition of ® (0, K) and S3, such that
for t e 2)(0, L + 2\X\am) one has
(60)
Fn(F{(t)) = F\(Fn(t))
e $(0, L) C %{B, K), n = 0, 1, 2, • • •
Now by (22)
(22')
A (t) = lim [Fn(t) - nam - ^ ^ - l o g n),
t<-%{Q,K).
For t€%{Q,L + 2\X\am) one has F{{t) in $){8,K) by (60) and from (60),
(22') one has
[17]
Permutable power series and regular iteration
A (Fx(t)) = Urn \Fn(F[(t)) - nam - ^
281
log n)
= lim \F{(Fn(t)) - nam - ^ ^ - log n),
whence by (49)
(61)
A{F{{t)) = A{t) + Xam.
Since the functions in (61) are regular the equation not only holds in
%{B, L + 2\X\am) but can be used in the form
(62)
Fx{t) = A_X{A (t) + Xam)
to give a regular continuation of P^t) throughout the simply connected
domain -<4_i(93i) of the £-plane.
We show that A^Qdj) D $(0, K + fydm). Let tx be a finite point on the
boundary of A^^Q^) and let s1 = A (t±) so that sx is on the boundary of S3X.
Put s2 = s1 — pdm (on the boundary of 93) and t2 = ^_!(s 2 ). The segment
F : s^ of length pdm joins the boundaries of 93 and 93i and its image A_X(T)
joins the boundaries of ®(0,X) = ^^(93) a n d ^ ^ ) . From (59) the length
of A_x(r) is less than twice the length of F and so is less than 2 pdm. Our
assertion follows.
Certainly then F{(t) is regular in <&(0,K + 2pdm) and
F{{®{e, K + 2 pdm)} C $(0, K) = ®.
Our K can be assumed to have been chosen so large that the succession of
transformations (37), (41), (45), (47) maps %{0,K) and ®(0,X + 2pdm) one
to one and conformally on to the domains ©(/, 0, K) and ®(j, 6,K-\-2pdm)
respectively, defined after (50). Since by lemma 4 the functions F*x{t)
and fx(z) correspond under these transformations we have proved that fx{z)
is regular in ©(/, 6,K + 2pdm) and that
(63)
fx{®(j, 6,K + 2pdm)} C ©(/, 0, K).
Such a result holds for any 1 ^ j ' ^ w s o that we may choose K so large that
(63) holds simultaneously for j = 1, 2, • • • m. Then if <&(d,K) is the star
(51) we have fx{z) regular in @(0, K + 2pdm) and
(64)
/A{<3(0, K + 2pdm)} C ©(0, K).
The choice of K depends in no way on the particular value of A chosen and
these results hold uniformly for all Xe&v.
We now remark that since X = 1 is in $„, so also is A = — 1. Replacing
M by
282
I. N. Baker
[18]
g(z) -= U(z) = z - am+1z<"-» + • • • = z + a'm+1z">+i +
in the above discussion and noting that gK = /_A SO that the class ®p is the
same for both / and g one sees that the only change is in the parameter y of
(41), (43), (50), which must now have such a value y' that
(43')
- a'm+1 e x p ( - iy'm) = am+1 e x p ( - imy') > 0
instead of
(43)
-«»,+i exp(-M»y) > 0.
It is legitimate to choose
y' =
The result (64) becomes in this case: There is a K > 0 such that for any
X e ®p the function gx(z) or /A(z) (since A e &p implies — X e $„) is regular in
the star <5'(d,K + 2pdm) and
(65)
/A{@'(0> -^
+ 2pdm)} C <S'{0, K).
In this result the star ®'{d,K) is (Jr®'(/. ^ . ^ ) precisely as in (50), (51).
Each domain ®'{j,d,K) is a domain bounded by a simple closed curve
analytic except at z = 0 where two arcs meet in the directions
arg*= — y H
(50')
2/ — ,
8
arg z = - y
71
(2/ - 2) - .
In fact @'( , K) has the appearance of ©(0, K) rotated through an angle of
— njm.
Fig. 4.
[19]
Pennutable power series and regular iteration
283
We choose a K valid both for (64) and (65). Since 0 < 0 < nj2, the set
@(0, K + 2pdm) u @'(0, K + 2pdm) u {z\z = 0}
includes a circle \z\ ^ p of positive radius. For every X e ®j, the function
fx(z) is regular in this circle and for \z\ ^ p one has by (64), (65)
/(*) € ©(0, K) u @'(0, if) u {z\z = 0},
which is a bounded set. The theorem is proved. The case m = 2, 0 = jr/4,
y = 0 is illustrated in Fig. 4. The boundary of @'(0, if) is dotted.
6. Convergence of the family fx
If ® is the set of X values corresponding to convergent members fx of the
family (7), our task is to describe the possible forms of S. Obviously ®
satisfies the lattice condition that it X e $£ and / j e t , then mX + njueSt for all
positive or negative integers m and n.
LEMMA
6. The set ft defined above is closed if X = oo be adjoined to it.
PROOF: Consider any sequence {Xn}, Xn e S, n = 1, 2, • • • which converges
to a finite value /i. All the Xn satisfy
for some p and by theorem 7 there is a circle |z| 5S p in which {/A } form a
uniformly bounded or normal family. We can therefore extract from {/A }
a subsequence, which we may as well assume to be {/A } itself, and which
converges uniformly to a limit g(z) in \z\ ^ p. g(z) is either the infinite constant or a regular function. But since /A (0) = 0 it follows that g(0) = 0
and in the same way
g(z) = * + A«m+1*m+i + • • • = /„(*),
since one has coefficientwise convergence also and the coefficients of fx(z)
are polynomials in X. Thus the function ffl(z) is convergent in \z\ 5S p and
From the lattice property of ® one sees that either every value of ® is a
point of accumulation of ® or no finite point of ® is a point of accumulation.
From this remark and lemma 6 follows
LEMMA 7. / / £ is any line through the origin of the complex X-plane then
® n&is either (i) the set {X = 0}, (ii) the set 8 or (iii) a set of the form {nX0},
n = 0, ± 1, ± 2 , • • • where Xo is one of the two values of least nonzero modulus
in Sn8.
284
I. N. Baker
[20]
8. If the set ® corresponding to the convergent members of a
family (7) contains a whole line £ which passes through the origin, then $ is
the whole X-plane.
THEOREM
In the proof of theorem 8 we use the following lemma.
8. If for the segment z = xe*e, ~ 1 ^ x ^ 1, the polynomial Pn(z)
of degree n satisfies
LEMMA
\Pn(z)\<M,
then for all complex z such that \z\ < 1 holds, one has
\Pn(z)\
It suffices to take the case 0 = 0. Then z — \{t + t'1) maps the
circumference of the unit ^-circle on to — 1 ^ x j£ 1 in the z-plane. Thus
PROOF:
(66)
\t»Pn{Ut + t-i)}\<M
for
|<| = 1.
Every non-real point z in \z\ < 1 has exactly one image point in \t\ < 1 and
this image point t has
\t\ > V2 - 1.
Thus for all z in \z\ ^ 1 one has from (66) and the assumptions of the lemma:
\Pn(z)\ < M(V2 + 1)-.
PROOF OF THEOREM 8.
Take p = 1. By theorem 7 there is a circle \z\ ^ p in which all /A(z) with
A e $ , are uniformly bounded by (say) M > 0. In particular this is the case
for all X eS n ® s . Hence the coefficients bn(X) in (7) satisfy
\bn[X)\ < MP~n
for
A € 2 n ft,, n = 2, 3, • • •
By lemma 8 it follows that
+ 1)-
for all
|A| rg p = 1, n = 2, 3, • • •
Hence (7) converges for \z\ < p / ( \ / 2 + 1) for all |A| £S p and ft, is the whole
circle |A| 5S £, so that ® is the whole plane.
A strengthening of theorem 8 is provided by
THEOREM 9: The set ® of theorem 8 has no finite points of accumulation
unless it consists of the whole plane.
PROOF. Take the case when $ is not the whole plane. We have to show
there is no finite point of accumulation in ®. By the lattice property of S it is
enough to show that A = 0 is not a point of accumulation. Suppose A = 0
is a point of accumulation. By theorem 8 any line £ passing through A = 0
[21]
Pennutable power series and regular iteration
286
will not belong wholly to ft and by lemma 7 there is a minimal \X(2)\ for
which / l ( 2 ) e 2 n S , 1^(8)1 > 0. This value may be infinite. By our assumption that 0 is a point of accumulation of ft it follows that there is a sequence
of lines 2(n), n — 1, 2, • • • for which X (£(»)) -*• 0. We may without loss of
generality assume that the lines £(») tend to a limit line 31. For any point
A2 e 9? and e > 0 we may find a line £(») of the sequence such that (a) the
perpendicular distance from Xx to £(») is less than e/2 and (b) |A(£(»))| <e/2.
It follows that there is a member of ft (on £(»)) closer to Xx than e. Thus Xx
is in the closed set ft and since it is an arbitrary pointof 9?,the whole line 31 is
contained in 2. By theorem 8 ft must be the whole plane. This contradiction
shows that X = 0 is not a point of accumulation and the theorem is proved.
10. If ft is the set of X corresponding to the convergent members
of the family (7) it has one of the forms
(i) the point X = 0,
(ii) {nXo}, n = 0, ± 1 , ± 2 , • • • with Xo^0,a
linear set,
(iii) {mX0 •+ nXJ, m = 0± ± 1, • • •, n = 0, ± 1, • • •, X0IXr not real,
Xo =£ 0x Xx # 0, a plane lattice, or
(iv) the whole plane.
THEOREM
PROOF: If (iv) does not apply the set ft is descrete. If ® contains points
other than X = 0 we take a finite circle \X\ ^ M containing such points.
\X\ ^ M contains at most a finite number of members of ft and we choose Xo
to be one of these with minimal modulus (excluding X = 0). Then alj points
nX0, n — 0, ± 1, ± 2, • • • lie in ft and if there are no others we have the
case (ii). Certainly these are the only members of ft which lie on the line £:
argA = argA0. If ft contains further members take that one of smallest
modulus which is first encountered on anticlockwise rotation of 2 and call it
Xx. Clearly Xo/X1 is not real, Xo =fc 0, Xx ^ 0, and by a standard argument ft is
e x h a u s t e d b y mX0 + nXx,m = 0, ± 1, ± 2, •••n = 0, ± 1, ± 2, • • •
Every such point is a point of ft and this is case (iv).
11. Cases (i), (ii) and (iv) of theorem 10 are realised.
We have not been able to find an example of case (iii).
THEOREM
PROOF, (i). The sequences {«„} and {Mn}, n = 2, 3, • • • are constructed
inductively as follows: a2 = M2 = 1 . If a2, • • •, an and M2, • • -Mn are
already constructed set
g(n, z) =z + a2z2 H
\- anzn
and let g\(n, z) be that series of the form
(67)
gx{n, z)=z
+ Xa2z* + £ cm(X)zm + bn+r{X)z^
TO-3
+ higher terms,
286
I. N. Baker
[22]
which commutes formally with g(n, z). Choose
(68)
M B+1
|A|£n
Then Mn+1 ^Mn.
Choose also
(69)
an+1 = Mn+1(n + 1)*«
Then the series
(70)
/iW = * + i«.*"
n-2
provides an example of case (1). In proof of this assertion: if
(71)
fx(z)=z
+ U2Z* + ZAn(X)z<>
3
is the commuting family of the form (7) containing (70), then calculation
shows that
For fixed X ^ 0 we have for sufficiently large n
An{X)
*B+i + K+itt) ^ WMn+1{n + 1J-+1 - Mn+1
Xan + bn(X)
\X\Mnti« + Mn
knn
by (69) and (68). This ratio tends to oo as n -> oo so that the series (71)
diverges for every X ^ 0. For A = 0 the series becomes identically fo(z) = z.
CASE (ii). In this part we consider the example provided by
and the family (7):
(72)
hiz) = * + |J*» + . . .
generated by it. In [1, p. 160] it was shown that the only real values in ® are
the integers. We show now that (72) cannot have a positive radius of convergence for any non-real X. Suppose X is not real and that (72) converges in a
circle \z\ < p, p > 0. In [1, p. 147 Satz 8] it was shown that the only nonconstant entire functions which commute with f^z) = e" — 1 are the natural
iterates /„ (?) ,n = 0, 1, 2, • • • Thus /A(z) is not entire and p < oo. The process
used in [1, pp. 161—2, Satz 17 parts (i), (ii) and (iii)] shows that fx{z)
has no singularity in the plane cut along the infinite segment
[23]
Pennutable power series and regular iteration
287
— oo < Re z ^ — p. There must be at least one singularity of fx(z) on the
circumference \z\ = p and this must occur at z = — p.
If on — p < x < 0 the function fx(z) takes any real value, say if /A(«I)
is real for — p <xx< 0, it follows from 0 > fx(x) > x,x < 0, that
/A(/I(*)) = /i(M*)) holds on [Xl, 0]. Then fMxJ)
= A ^ f o ) ) is a real
quantity. Thus if x% = f{xj) we have 0 > x2 > xx and fx(x2) real. Defining
zn = /(«„_!) for » = 2, 3, • • • we have 0 > a?n > «„_! and fx(xn) real. There
is a number I such that * B - > / ^ 0 a s » - ^ - o o and one must have fx(l) = /
so that / = 0. But since
Xx*
fx(x) = x+—-\
, A not real,
z
it is impossible that /*(»„) be real for xn -> 0. Consequently fx(z) takes no real
value in (— p, 0).
Since for x < 0 one has fx(x) > x, it is possible to choose px> p such that
/ i ( ~ Pi) > — P- Then /A(/I(«) is regular for — px < x < 0 and takes no real
values there. Therefore f-i{fx(fi(x))) is regular on — px < x < 0 and gives
a continuation of /A(«) there, which is regular over z = — p. This contradicts
the fact that z = — p is a singularity of /*(•?), so that the assumption p > 0
is false.
CASE
(iv). This is realised by the family
X — AZ
n_o
which forms a continuous group of substitutions.
7. Existence of an infinitesimal transformation
In case (iv) of theorems 10 and 11 one has a local continuous group of
functions fx(z) with complex parameter X and
The individual transformations have expansions of the form (7). If we restrict
X to \X\ ^ 1, there is a 6 > 0 such that all these fx(z) converge in \z\ ^ d,
where they also satisfy \f(z)\ < M for a suitable constant M. The coefficients
in (7) satisfy
\bn(X)\ ^ MS-"
for
\X\ ^ 1.
Now for n ^ m + 1 we have shown in theorem 2 that bn{X) is polynomial
degree at most (» — m). From (12) bn(0) = 0. By the lemma of Schwarz
288
[24]
I. N. Baker
^ M8-"
for
\X\ <L
\b'm{0)\ ^
Md-
and
Also
is a polynomial in X whose constant term vanishes.
\pn(X)\ ^ \bn(X)\ + \b'n(0)\ ^ 2M8-*
for
\X\
and since pn(0) = 0 one has
(73)
\pn(X)\ =g 2Ma-»|A| for
\X\ ^ 1.
The series
n—m+2
converges for |*| < 8, where moreover
//A(*] — *\
00
-
n—m+1
^**
oo
1/1*1"^
\X\
3m+1
from (73). Thus
(75)
/(*)
lim
2M\z\m+2
{8
"
lZl)
A
A-M)
X
holds for |*| < 8 and/(*) is the infinitesimal transformation of the group (7).
The function /,,(*) may be substituted in place of * in (75), provided we make
8 smaller, and (75) becomes
(76)
'(/,(*)) =
The existence of the derivative (76) shows that in a region of the form
|*| < d, \/t\ < p the function /,,(*) is analytic in both //, and *. Taking a
fixed * ^ 0 in |*| < 8 and letting //, grow from 0 to A along the path p = tX,
0 ^ t ^ 1 we find that f^z) traces a curve F in |*| < 8. If 8 and A are
sufficiently small this curve F is simple and does not enclose the origin.
Integrating (76) along this curve
[25]
Permutable power series and regular iteration
289
with the boundary conditions fM(z) = z when /u = 0 and fx(z) when p = X
we obtain
(77)
where
i{z) =
(78)
r* dz
r r r has an expansion of the form
J J\z)
J\z)
Ax{z) = 2 «.*-" + c0 log * + I cnz\
n=l
Such a formula (77) is of the same type as (23) and (61). A function Ax{z)
satisfying (77) is called an Abel function and the equation itself Abel's
functional equation. We see that there is an Abel function intimately connected with the infinitesimal transformation/^) of the group (7) and that this
functional equation can be used to define fx(z) at least in partial neighbourhoods of z = 0. On the other hand the logarithmic term in (78) will in general
be present and the Abel function will not be single valued around z = 0.
In general then one can not expect to study the continuous iteration near
a fixpoint of multiplier 1 by finding an Abel function regular up to poles
at z = 0. A case where the function Ax(z) is, however, single valued at z = 0
is provided by the family
for which I(z) = z2 and Ax{z) = — z~1.
One may remark that from an Abel function of the form (78) or an infinitessimal transformation (74) one can construct a group (7). If for example the
convergent series (74) is given, one has only to solve the equation (76),
that is the differential equation
(79)
£ = /(»)
with boundary condition w = z at p = 0. The solution w(/i, z) is analytic
in both n and z in a neighbourhood of n = 0, z = 0 (e.g. Bieberbach [3]).
The function w(fi, z) has an expansion in terms of z of the form
w{/*,z)=z
+ fiam+1zm+1 -j
We put
Xi = »{p. »(A. *)}
Xi = w{ji + X, z).
290
I. N. Baker
[26]
Then dxJSfi = I{xi) and <ty2/3/M = I{xz)- These are simply equation (79)
and are to be integrated with the boundary condition xx = X2 = w&, z)
when p = 0. Thus x\ = Xz anc^ the functions w(p, z) form a commuting
group of series (7) whose infinitesimal transformation is I(z).
In the cases (i), (ii) and (iii) where the family (7) does not converge for
all X, one may form the series (74) corresponding to I(z) but the work of the
preceding part of this paragraph shows that the series will not converge.
For convergent series I(z) correspond to case (iv) only. In the cases (ii)
and (iii) it may be shown by further analysis that there is indeed a family of
continuous iterates fx(z) satisfying /*(/,,) = fpifx) but most of these have a
singularity at the origin. G. Szekeres [15] has investigated such situations.
8. Some applications to permutable entire functions
In § 2, theorem 6, a simple application of theorems 1—5 was made to the
theory of entire functions. In the second example of theorem 11 the entire
function e* — 1 occurred. It is natural then to investigate the case when
certain members of a commuting family (7) are entire functions. We prove
first the following
z
THEOREM 12. If {fx( )} is a commuting family (7), then the numbers X such
that fx(z) is entire form a discrete and hence countable set.
LEMMA 9. / / the series
/
(z\
—• Z
I
(L
Z™~^~^~ -4—
:•
(L Z9*
n—m+2
has a positive radius of convergence, then for given 0 with 0 < 0 < n/2 and all
sufficiently large K there is a star ©(0, K) of the form (51) such that
(80)
(81)
/ 1 {©(0,iiC)}C©(0, J K:) >
/„(*) -> 0 in ©(0, K) for all z e ©(0, K).
Further there is a star ©'(0, K) of the form (50'), (51) (cf. proof of theorem 7)
such that
(82)
(83)
U{&(
f-n{z) -> 0 in @'(0, K) for all z e ©'(0, K).
PROOF: This follows immediately from the last part of lemma 4 with
A = 1 together with the fact that, in the notation of lemma 4:
F z
n{ ) -*• °o a s »->«> for any ze'S)(d,K) by (21).
We remark that since K may be taken arbitrarily large, the regions ©(0, K)
and ©'(0, K) may be assumed interior to any given circle centred at z = 0.
[27]
Permutable power series and regular iteration
291
PROOF OF THEOREM 12. There is nothing to prove except when case (iv)
of theorem 10 applies — and when for some X = 0 the function fx(z) of (7) is
entire. There is no loss of generality in assuming that for X = 1 the function
fx(z) is entire. Applying lemma 9 to fx(z) we form © = ©(6, K) and ©' =
©'(0, K) and note that © u ©' may be supposed interior to a given circle
\z\ < t,t > 0. According to theorem 7 there is to any p > 1 a constant p > 0
and a constant M > 0 such that for all \X\ < p the series fx(z) converges in
\z\ ^ p and |/A(z)| ^ Af for \z\ ^ p. Then
|/A(*)|=S—
for
!*!<£/,.
Choose the above t so that 0 < t 5S /» and so that for |^| < t and |A| ^ p,
the value /A(z) is contained in a region where f(z) is univalent. This is possible
because \fx(z)\ < Mtjp.
Now, as was pointed out in the poof of theorem 7 (cf. Fig. 4), the domain
© u ©' contains a circle \z\ < a, a > 0. Since fx(z) = z + am+1zm+1 + • • •
holds, the sequence {fn(z)} is not normal near 2 = 0. From Fatou [6], [5],
z = 0 is a point of accumulation of fixpoints of iterates fn(z). Thus there
exist a point £ ^ 0 and an integer N > 1 such that
(84)
|£|<«r,
£«@u@',
M*)=f.
/,(£)#*
for / < iV.
Now for » = 1, 2, • • • one has /„(£) = /,(£) where 1 ^ / ^ iV and / = n
(modulo N). Moreover f}(!;) =£ fk(g) for 1 :g / < h ^ N. Thus the sequence
{/„(£)} does not converge to 0. By (80) and (81) f 4 ©. Thus by (84) f e ©'.
Suppose that there is an infinity of values X in \X\ ^ p for which /A(z) is
entire. These values have a point of accumulation (i, \[i\ fS p. There is a
sequence of values Xn, n = 1, 2, • • • in which Xm ^ Xn for m =£ n, such that
Xn -*• ft and fx (I) ->//,(£) as » -»• oo. The values A (|) are fixpoints of fN[z)
and thus members of a discrete set. Hence from a certain « onwards all
a r e ne
/A,(£)
* same. Thus if a = Xn and /? = AB+1 9^ a are suitable members of
the sequence we have / „ ( ? ) = fp{£)Define inductively fx = £, | n + 1 = /_i(l n ) where /_ t may be taken to be
the series expansion f^z) = z — &m+\Zm+1 + • • • and by lemma 9 the
uniquely defined points fB lie in ©' for « = 1, 2, 3, • • • As n ->- 00 we
have £„ -> 0. We assert that /„(!„) = ffi[Sn) for all n. This is certainly true
for n = 1. Suppose the result holds for » = 1, 2, • • • ,m — 1. Then
/itf.fc.)} = LV(U)
= /.(f«-i) = /,(£«-i) = /i{/ 4 (fJ}-
But / a (| m ) and /^(fm) lie in a region where /(z) is univalent and so must be
equal. Our assertion is proved by induction.
Since / a (f n ) = //£„) and £„ ->• 0 as n -> c» it follows that /a(z) = /^(z)
which can only be true if Xn = a = ^ = AB+1. This is a contradiction and it
292
I. N. Baker
[28]
follows that our hypothesis of an infinite set of X with \k\ ^ p and /A entire
must be false. Since this applies for every p > 1 the proof of theorem 12 is
complete.
We have also shown that the entire members of a family (7) form at most
a countable set. This set always contains the linear polynomial fo(z) = z and
no other linear polynomial. There may be also a polynomial of degree two
or more among the functions /A(z) and in this case the family can include
no transcendantal entire function. Conversely if there is a transcendental
member of the family (7) then the only polynomial member is fo(z) = z.
These assertions follow from a result of Baker [1, p. 141] and Ganapathy
Iyer [8]: no entire transcendental function can commute with a polynomial
of degree two or more.
We now combine theorems 6 and 12 to obtain
THEOREM 13. / / F(z) is a non-linear entire junction exactly one of whose
fixpoints has multiplier 1, then the entire functions permutable with F(z)
form a countable set.
PROOF: AS in theorem 6 we may assume the fixpoint to be z = 0 so that
F(z) has an expansion
F(z) = z = an+1z>»+i +
f
anz\ m ^ 1, am+1 * 0.
n-m+2
The constant functions permutable with F(z) are the solutions c of F(c) = c
which form at most a countable set. The non-constant functions permutable
with F(z) are either members of the family (7): Fx(z) of theorem 2 or by
theorem 4 they have the form H o Fx{z) where Hm(z) = z. The entire functions of the form Fx(z) form at most a countable set. If H o F\{z) is entire
then so is its w-th interate FmX(z) which is in (7). Thus the entire functions
of this second kind are also at most countable and the theorem is proved.
One may ask whether the set of entire functions permutable with a given
entire function is countable under more general hypotheses then those of
theorem 13. This is indeed so but other methods of proof are required.
Finally we prove
THEOREM
14. / / the members
ft(z) =z + an+1zm+i + f
bn{l)zn, « m+ i * 0, m ^ 1,
and
fx(z) = z + Aa m+1 * m « + f bn{X)z»
m+2
of the family (7) are entire, then X > 0.
[29]
Pennutable power series and regular iteration
293
Suppose X = leta, I > 0, a ^ O , — n < a. ^.TZ. Choose 0 with
0 < 20 < |a|. By lemma 9 there is a star © = <S>(d,K) of the form (51)
such that ^((S) C © and fn(z) -+ 0 in © for z e ©. By (50) the m branches of
the star © lie one in each of the angles
PROOF:
(86)
_ y + 1 - (2; - 3)£ < arg* < - y - I - (2/ - 1)£,
7 = 1,2, ••-,»»,
i.e. in angles of magnitude 2{n — 6)/m spaced evenly around z = 0 with a
gap of 20/m between neighbouring angles. The boundary of © is tangent at
z = 0 to the arms of the angles (85). By (43) the term y in (85) is determined
by — am+1 exp (— iym) > 0.
Applying the same reasoning to the function /A(z) we see that there is a
star 9 = #(0, K) of the form (51) such that /A(Qf) C 3 and fnX{z) -> 0 in Qf
for z e 8. The »* branches of 9 lie one in each of the angles
(86)
_
/ +
l_ ( 2 /_3 ) £ < arg*<-/-^-(2/-l)^,
/ = 1, 2, • • • tn,
where by (43) y' is determined by
— Xam+r exp(— »>'») > 0.
Since X = leia, I > 0, we may put / = y + «/m. Thus by (85), (86) the
appearance of Qf is that of © rotated about the origin through an angle
— <x./m. Since 2d/m < \ct\ftn < 2(n — B)\m, it follows that © u Qf contains a
circle \z\ < r,r > 0. We remark further that since the K in ©(0,i£) and
3f(0, K) may be arbitrarily enlarged, the region © u 9 may be assumed to
be contained in an arbitrarily small circle centred at the origin.
By the results of Baker [2] or Rosenbloom [13], since fx(z) is a nonlinear
entire function, it follows that for all but at most one of the values / = 2,
3, 4 the equation ft(z) = z has a solution g which is not a solution oifk(z) = z
for 1 5S k < /. Thus there are points £ and rj with /(£) ^ £, f(r/) =fiif, g ^r/,
/i2^) = £> fiziv) — V- Clearly £ ^ 0, r\ ^ 0. By the above remark we may
suppose © u Of so close to z = 0 that f ^ S u Q f , jy^©uOf. Now since
/-AW = * + «^m+i^ m+1 + • • • . » = 1, 2, •••
it follows that the sequence {/^J is not normal in \z\ < r and there is a value
tal < r and an integer JV ^ 1 such that
) = f or
294
I. N. Baker
[30]
Suppose firx{zi) — £> which is no loss of generality. Then since /12(£) = £,
we have for any positive integer m
fl2mifN^l)} = f, »f = 1, 2, S, • • •
If z1 e © then /12m(^i) -> 0 as m -*• oo and hence /jvA{/i2m(zi)} -*• 0- Therefore
^ ^ ©, zx e Qf. But then fyxfa) e Of for all N = 1, 2, • • •, which contradicts
fNx(zi) — f ^ ® u ©. Thus the assumption a ^ 0 is false.
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[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
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